An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is xm. The projectile breaks into two pieces which fly apart horizontally. The larger piece, m3, has three times the mass of the smaller piece, m2. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station. Neglect air resistance and effects due to the earth's curvature.
How far away, xf, from the original launching point does the larger piece land? Express your answer in terms of some or all of the given variables m1, xm, and g (enter m_1 for m1 and x_m for xm).
xf=
To find the distance xf at which the larger piece lands, we need to consider the conservation of linear momentum.
Before the explosion, the total momentum of the system is zero because the projectile is at the top of its trajectory and has zero horizontal velocity. Therefore, after the explosion, the total momentum of the system must still be zero.
Let's denote the initial velocity of the smaller piece as v1, and the initial velocity of the larger piece as v2. Since the smaller piece returns to the launching station, its final velocity vf1 would be equal to -v1 (opposite direction).
Using the conservation of momentum, we have:
m2 * vf1 + m3 * vf2 = 0
Since m3 = 3 * m2, we can express the equation as:
m2 * (-v1) + 3 * m2 * vf2 = 0
Simplifying and rearranging, we get:
vf2 = (1/3) * v1
Now, let's consider the motion of the larger piece. Since there is no horizontal force acting on it (neglecting air resistance), its horizontal speed remains constant throughout its motion. We can use the equation:
xf = xm + vt
where xf is the final distance, xm is the horizontal distance between the launch point and the explosion, v is the constant horizontal velocity, and t is the time it takes for the larger piece to reach the ground.
To find t, we can use the kinematic equation:
y = y0 + v0y*t + (1/2) * a * t^2
Since the larger piece is moving horizontally, its initial vertical velocity v0y and vertical acceleration a are zero. Therefore, the equation becomes:
0 = 0 + 0 + (1/2) * g * t^2
Simplifying, we get:
t^2 = 0
This shows that the time required for the larger piece to reach the ground is 0 seconds. Therefore, the larger piece lands at the same time as the explosion, directly below the point of the explosion.
Substituting t = 0 in the equation xf = xm + vt, we get:
xf = xm
Hence, the larger piece lands at the same horizontal distance xm from the original launching point. Therefore, the distance xf is equal to xm.